Friday, December 24, 2010
Lo, how a Rose e'er bloomin'
1. Es ist ein Ros' entsprungen (adap. Praetorius 1609)
2. Nu är det jul (Trad.)Ale Möller (Mandola and flute)Lisa Rydberg and Esbjörn Hazelius (Violins)Roger Tallroth (Bouzouki)Olle Linder (Double Bass)A Blessed Christmas to all in this sainted year 2010!
Sunday, December 19, 2010
Sjömansvisan
My first complete transcription on Sibelius 5, a slow-ish sailor's ballad:
Title: Sjömansvisan (The Sailor's Song)Lyrics: Traditional Swedish
Music: Ulrika Bodén
As performed by: Ranarim (Sweden)
Code: T01.se / [see complete PDF]
Lyrics:
Skeppet seglar och vinden blåser
nu hissar vi våra ankar
vi ska segla till vännen kära
den vi har i våra tankar
Räck mig nu den snövita handen
om du vill bliva min kära
och vi ska tacka både fader och mor
om vi varandra skall äga
Dagen lider och solen skrider
och stjärnorna tändas upp igen
inte kan jag till vännen komma
varken åka, rida eller gå
Räck mig nu den snövita handen
om du vill bliva min kära
men jag får vänta till sol och sommardag
då jag seglar över böljan blå
Translation:
The vessel sails and the wind blows
Now we raise our anchors
We’ll sail to our beloved,
who is in our thoughts
Reach out your snow-white hand
If you will be my love
We shall thank both Mother and Father
if we shall have each other
The day passes and the sun moves across the sky
The stars are again lit
I cannot reach my beloved,
neither ride nor wander
Reach out your snow-white hand
If you will be my love
But I must wait for sun and summer days
to sail the bounding main
Thursday, December 02, 2010
The Aranjuez Diaries
Concierto de Aranjuez 3rd movement: Allegro Gentile
Joaquín Rodrigo (Comp.) / Narciso Yepes (12-stringer)
Before each Wednesday or Friday evening lecture, Prof Leung liked to play Concierto de Aranjuez from Youtube as a kind of mood setting. Usually only movements 1 or 2 were played, but the third movement is my favourite, so here you go.
List of Solutions in GEK1517 Disclaimer: These are my own solutions, not the ones derived in class; so if I couldn't get a solution or the lecturer's solution is the same as mine, I would not post it. It should follow that the demonstrations are imperfect and not final.Tutorial 1: Black Fridays
Tutorial 3: Cracking Tiles
Tutorial 6: Fibonacci's Rabbits
Tutorial 6: Turtle Crawl
Tutorial 8: Szymborska's Utopia
Examination: Proving the Obvious
Miscellany: Application to MA1505: The Nabla Vector (use of metaphors)
Application to MA1505: Theorems of Vector Calculus (graphical demonstration)
Quote: Bishop George Berkeley (how not to learn science)
Quotes: Polanyi, Annan and Byers (somewhat random)
Concierto de Aranjuez, second movement
About the Course A course in the ideas behind breakthroughs in mathematics. In the course, complicated ideas are broken down into simple ones and presented as the mathematicians would have imagined it. This idea is mirrored in the test/exam/tutorials, where one arrives at proofs using guided instructions. This course reveals the joy of mathematics. Despite this, it probably put off some people from maths forever.
Because some things cannot be unlearnt, the ideas of GEK1517 has stuck on me and follows me wherever I go, surfacing occasionally and unexpectedly in my other modules (Calculus, physics, chemistry and even writing!).
Proving the Obvious
GEK1517 Examination / Question 1:
(i) Preface: Peano's Axioms
UT: The set of all natural numbers ℕ = {1, 2, 3, ...}
UT: 1
UT: Successor (here denoted by ')
(A1): 1 is a natural number
(A2): If a number a is a natural number, its successor is also a natural number
(A3): 1 is not the successor to any natural number
(A4): If two numbers have the same successor, they are the same number.
(A5): If 1 is a member of set K, and if for every natural number a, a is a member of K implies that a' is a member of K, then K is the set of all natural numbers. (Mathematical Induction).
The proofs in the following also feature reflexive equality (Ar) strongly. This is the first axiom in Peano's original list of axioms but was omitted from class notes. Essentially, it means x = x.
(ii) In the Peano axiomatic system for natural numbers, addition '+' is given inductively by the following two properties.
(+1): a + 1 = a' for a ∈ ℕ
(+2): a + b' = (a + b)' for a, b ∈ ℕ
Using (+1), (+2), and mathematical induction, show that
1 + a = a + 1 for all a ∈ ℕ
Solution:
Let P(a) denote (1 + a = a + 1 for all a ∈ ℕ)
P(1): 1 + 1 = 1 + 1 (true because (Ar))
P(k): 1 + k = k + 1 (force true)
P(k'): 1 + k' = k' + 1
(+1): 1 + k' = (k + 1) + 1
(+1): 1 + k' = (k + 1)'
(+2): (1 + k)' = (k + 1)'
(A4): 1 + k = k + 1 (true because P(k))
(A5): Statement proven for for all a ∈ ℕ
Lemma 1 (L1): Commutativity of +
To prove: a + b = b + a for all a, b ∈ ℕ
Let P(a) denote (a + b = b + a) , b ∈ ℕ
P(1): 1 + b = b + 1 (true because (ii))
P(k): k + b = b + k (force true)
P(k'): k' + b = b + k'
(+1): k + 1 + b = b + k'
(ii): k + b + 1 = b + k'
(+1): k + b' = b + k'
(+2): (k + b)' = (b + k)'
(A4): k + b = b + k (true because P(k))
(A5): Statement (L1) proven for for all a, b ∈ ℕ
(iii) In the Peano axiomatic system for natural numbers, multiplication '×' is given inductively by the following two properties.
(×1): a × 1 = a for a ∈ ℕ
(×2): a × b' = (a × b) + a for a, b ∈ ℕ
Using (×1), (×2), and mathematical induction, show that
a × 2 = a + a, and
a' × b = a × b + b for a, b ∈ ℕ
Solution part 1:
Let P(a) denote (a x 2 = a + a)
P(1): 1 × 2 = 1 + 1
What is 2? Take 2 = 1' (oh no...)
P(1): 1 × 1' = 1 + 1
(×2): (1 x 1) + 1 = 1 + 1
(×1): 1 + 1 = 1 + 1 (P(1) true because (Ar))
P(k): k × 2 = k + k (force true)
P(k'): k' × 2 = k' + k'
(2 = 1'): k' × 1' = k' + k'
(×2): (k' × 1) + k' = k' + k'
(×1): k' + k' = k' + k' (true because (Ar))
(A5): Statement proven for for all a ∈ ℕ
Solution part 2: Commutativity of ×
Let P(b) denote (a' × b = (a × b) + b), a ∈ ℕ
P(1): a' × 1 = (a × 1) + 1
(×1): a' = a + 1
(+1): a' = a' (true because (Ar))
P(k): a' × k = (a × k) + k (force true)
P(k'): a' × k' = (a × k') + k'
(×2): (a' × k) + a' = (a × k) + a + k'
using P(k): (a × k) + k + a' = (a × k) + a + k'
(+2): (a × k) + (k + a)' = (a × k) + (a + k)'
(L1): a + k = k + a
so (a × k) + (a + k)' = (a × k) + (a + k)' (true because (Ar))
(A5): Statement proven for for all a, b ∈ ℕ
Colophon:
The idea: you must forget all your arithmetic before you tackle this question. There is no arguing that "everyone knows" a+b = b+a; the only way is to prove it from the given axioms and properties. This is the intellectual equivalent of bailing out from a plane into the rainforest and building civilisation from the bottom up, right down to foraging for food, starting a fire, etc.
This is the first and only real proof in the series... I hope this is rigorous enough, and I hope the formatting makes it clear to follow (Facebook users please allow yourselves to be redirected to the "original post")
Regrettably, I did this only today, when the dust has settled over the exam period and freedom, merriment and general goofing off is in order. Never mind, enjoy the holidays folks.
(i) Preface: Peano's Axioms
UT: The set of all natural numbers ℕ = {1, 2, 3, ...}
UT: 1
UT: Successor (here denoted by ')
(A1): 1 is a natural number
(A2): If a number a is a natural number, its successor is also a natural number
(A3): 1 is not the successor to any natural number
(A4): If two numbers have the same successor, they are the same number.
(A5): If 1 is a member of set K, and if for every natural number a, a is a member of K implies that a' is a member of K, then K is the set of all natural numbers. (Mathematical Induction).
The proofs in the following also feature reflexive equality (Ar) strongly. This is the first axiom in Peano's original list of axioms but was omitted from class notes. Essentially, it means x = x.
(ii) In the Peano axiomatic system for natural numbers, addition '+' is given inductively by the following two properties.
(+1): a + 1 = a' for a ∈ ℕ
(+2): a + b' = (a + b)' for a, b ∈ ℕ
Using (+1), (+2), and mathematical induction, show that
1 + a = a + 1 for all a ∈ ℕ
Solution:
Let P(a) denote (1 + a = a + 1 for all a ∈ ℕ)
P(1): 1 + 1 = 1 + 1 (true because (Ar))
P(k): 1 + k = k + 1 (force true)
P(k'): 1 + k' = k' + 1
(+1): 1 + k' = (k + 1) + 1
(+1): 1 + k' = (k + 1)'
(+2): (1 + k)' = (k + 1)'
(A4): 1 + k = k + 1 (true because P(k))
(A5): Statement proven for for all a ∈ ℕ
Lemma 1 (L1): Commutativity of +
To prove: a + b = b + a for all a, b ∈ ℕ
Let P(a) denote (a + b = b + a) , b ∈ ℕ
P(1): 1 + b = b + 1 (true because (ii))
P(k): k + b = b + k (force true)
P(k'): k' + b = b + k'
(+1): k + 1 + b = b + k'
(ii): k + b + 1 = b + k'
(+1): k + b' = b + k'
(+2): (k + b)' = (b + k)'
(A4): k + b = b + k (true because P(k))
(A5): Statement (L1) proven for for all a, b ∈ ℕ
(iii) In the Peano axiomatic system for natural numbers, multiplication '×' is given inductively by the following two properties.
(×1): a × 1 = a for a ∈ ℕ
(×2): a × b' = (a × b) + a for a, b ∈ ℕ
Using (×1), (×2), and mathematical induction, show that
a × 2 = a + a, and
a' × b = a × b + b for a, b ∈ ℕ
Solution part 1:
Let P(a) denote (a x 2 = a + a)
P(1): 1 × 2 = 1 + 1
What is 2? Take 2 = 1' (oh no...)
P(1): 1 × 1' = 1 + 1
(×2): (1 x 1) + 1 = 1 + 1
(×1): 1 + 1 = 1 + 1 (P(1) true because (Ar))
P(k): k × 2 = k + k (force true)
P(k'): k' × 2 = k' + k'
(2 = 1'): k' × 1' = k' + k'
(×2): (k' × 1) + k' = k' + k'
(×1): k' + k' = k' + k' (true because (Ar))
(A5): Statement proven for for all a ∈ ℕ
Solution part 2: Commutativity of ×
Let P(b) denote (a' × b = (a × b) + b), a ∈ ℕ
P(1): a' × 1 = (a × 1) + 1
(×1): a' = a + 1
(+1): a' = a' (true because (Ar))
P(k): a' × k = (a × k) + k (force true)
P(k'): a' × k' = (a × k') + k'
(×2): (a' × k) + a' = (a × k) + a + k'
using P(k): (a × k) + k + a' = (a × k) + a + k'
(+2): (a × k) + (k + a)' = (a × k) + (a + k)'
(L1): a + k = k + a
so (a × k) + (a + k)' = (a × k) + (a + k)' (true because (Ar))
(A5): Statement proven for for all a, b ∈ ℕ
Colophon:
The idea: you must forget all your arithmetic before you tackle this question. There is no arguing that "everyone knows" a+b = b+a; the only way is to prove it from the given axioms and properties. This is the intellectual equivalent of bailing out from a plane into the rainforest and building civilisation from the bottom up, right down to foraging for food, starting a fire, etc.
This is the first and only real proof in the series... I hope this is rigorous enough, and I hope the formatting makes it clear to follow (Facebook users please allow yourselves to be redirected to the "original post")
Regrettably, I did this only today, when the dust has settled over the exam period and freedom, merriment and general goofing off is in order. Never mind, enjoy the holidays folks.
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